Abstract Nonsense

Crushing one theorem at a time

Necessary and Sufficient Conditions for Isomorphism

Point of post: This post will give a nice result which will give us necessary and sufficient conditions for a map between vector spaces to be an isomorphism. I will cite this post every now and then when I would like to shorten a proof.

Theorem: Let \mathscr{U} and \mathscr{V} be finite-dimensional vector spaces over the field F. Then, a linear mapping f:\mathscr{U}\to\mathscr{V} is an isomorphism if and only if there exists bases \{x_1,\cdots,x_n\}=\mathcal{B}_1 for \mathscr{U} and \{y_1,\cdots,y_m\}=\mathcal{B}_2 for \mathscr{V} such that


is a bijection.

Proof: Suppose first that there existed bases \mathcal{B}_1 and \mathcal{B}_2 such that the above is true and assume without loss of generality that f(x_j)=y_j. To prove injectivity suppose that f(x_1)=f(x_2). Then, since \mathcal{B}_1 is a basis for \mathscr{U} we know that \displaystyle x_1=\sum_{j=1}^{n}\alpha_j x_j and \displaystyle x_2=\sum_{j=1}^{n}\beta_j x_j for some \alpha_1,\cdots,\alpha_n,\beta_1,\cdots,\beta_n\in F. Thus, by the linearity of f we have that

\displaystyle \sum_{j=1}^{n}\alpha_j f(x_j)=f(x_1)=f(x_2)=\sum_{j=1}^{n}\beta_j f(x_j)

But, this says that

\displaystyle \sum_{j=1}^{n}\alpha_j y_j=\sum_{j=1}^{n}\beta_j y_j

and since \mathcal{B}_2 is a basis for \mathscr{V} it follows that \alpha_j=\beta_j,\text{ }j=1,\cdots,n and so x_1=x_2. Now, to prove surjectivity we merely note that for any y\in\mathscr{V} we have that there is some \beta_1,\cdots,\beta_m\in\mathcal{B}_2 such that \displaystyle y=\sum_{j=1}^{n}\beta_j y_j. But, it follows that

\displaystyle y=\sum_{j=1}^{n}\beta_j y_j=\sum_{j=1}^{n}\beta_j f(x_j)=f\left(\sum_{j=1}^{n}\beta_j x_j\right)

from where surjectivity follows. Thus, f is a bijective linear mapping and thus an isomorphism.

Conversely, it is trivial that if f is an isomorphism that the image under any basis for \mathscr{U} is a basis for \mathscr{V}, and since the restriction of a bijection is a bijection the conclusion follows. \blacksquare


November 1, 2010 - Posted by | Algebra, Linear Algebra | , , ,

1 Comment »

  1. […] linearity. By a previous post we may conclude that is an isomorphism. Also, we note that if and we see […]

    Pingback by Halmos Section 24 and 25: Tensor Product and Product Bases « Abstract Nonsense | November 1, 2010 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: